Answer
The solutions are $x=-\dfrac{1}{2}\pm\dfrac{\sqrt{21}}{2}$
Work Step by Step
$\dfrac{1}{x^{2}-3x+2}=\dfrac{1}{x+2}+\dfrac{5}{x^{2}-4}$
Factor all the rational expressions present in the equation completely:
$\dfrac{1}{(x-1)(x-2)}=\dfrac{1}{x+2}+\dfrac{5}{(x-2)(x+2)}$
Multiply the whole equation by $(x-1)(x-2)(x+2)$:
$(x-1)(x-2)(x+2)\Big[\dfrac{1}{(x-1)(x-2)}=\dfrac{1}{x+2}+\dfrac{5}{(x-2)(x+2)}\Big]$
$x+2=(x-1)(x-2)+5(x-1)$
$x+2=x^{2}-2x-x+2+5x-5$
Take all terms to the right side and simplify:
$0=x^{2}-2x-x+2+5x-5-x-2$
$0=x^{2}+x-5$
Rearrange:
$x^{2}+x-5=0$
Use the quadratic formula to solve this equation. The formula is $x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
In this case, $a=1$, $b=1$ and $c=-5$
Substitute the known values into the formula and evaluate:
$x=\dfrac{-1\pm\sqrt{1^{2}-4(1)(-5)}}{2(1)}=\dfrac{-1\pm\sqrt{1+20}}{2}=...$
$...=\dfrac{-1\pm\sqrt{21}}{2}=-\dfrac{1}{2}\pm\dfrac{\sqrt{21}}{2}$
